Topic 9 The Mole
I. Molar Conversions
I
II
III
IV
A. What is the Mole?
A counting number (like a dozen)
 Avogadro’s number (NA)
 1 mol = 6.02  1023 representative
particles (atoms, ions, formula units,
or molecules

If you could spend 1 million dollars per
second, how many years will it take to
spend a mole of dollars, given there are
31536000 seconds per year.
1.91 x 1010 years or 19.1 billion years
A. What is the Mole?

1 mole of hockey pucks would
equal the mass of the moon!

1 mole of basketballs would fill a
bag the size of the earth!

1 mole of pennies would cover
the Earth 1/4 mile deep!
A. Moles
Used as a Conversion Factor:
 1 mole = 6.02 x 1023 so…
 Moles converted to Particles =

# mol 6.02  1023 Particles
1 mol
A. Moles

Rep. Particles converted to moles =
# particles 1 mol
6.02  1023 Particles
B. Molar Mass

Mass of 1 mole of an element or
compound.

Atomic mass tells the...
 atomic mass units per atom (amu)
 grams per mole (g/mol)

Round to 2 decimal places
B. Molar Mass Examples

carbon
12.01 g/mol

aluminum
26.98 g/mol

zinc
65.39 g/mol
B. Molar Mass Examples

water
 H2O
 2(1.01) + 16.00 = 18.02 g/mol

sodium chloride
 NaCl
 22.99 + 35.45 = 58.44 g/mol
B. Molar Mass Examples

sodium hydrogen carbonate
 NaHCO3
 22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol

sucrose
 C12H22O11
 12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
C. Molar Conversions
molar
mass
6.02  1023
MASS
NUMBER
MOLES
IN
GRAMS
OF
PARTICLES
(g/mol)
(particles/mol)
C. Molar Conversion Examples
 How
many molecules are in
2.50 moles of C12H22O11?
6.02  1023
2.50 mol molecules
1 mol
= 1.51  1024
molecules
C12H22O11
C. Molar Conversion Examples
 How
many moles of carbon are
in 26 g of carbon?
26 g C 1 mol C
12.01 g C
= 2.2 mol C
C. Molar Conversion Examples
the mass of 2.1  1024
molecules of NaHCO3.
 Find
2.1  1024
molecules
1 mol
84.01 g
6.02  1023 1 mol
molecules
= 290 g NaHCO3
Topic 9 – The Mole
II. Formula
Calculations
I
II
III
IV
A. Percentage Composition
 the
percentage by mass of each
element in a compound
mass of element
% composition 
 100
total mass
A. Percentage Composition
 Find
%Cu =
%S =
the % composition of Cu2S.
127.10 g Cu
 100 =
159.17 g Cu2S
79.852% Cu
32.07 g S
159.17 g Cu2S
 100 =
20.15% S
A. Percentage Composition
 Find
the percentage composition
of a sample that is 28 g Fe and
8.0 g O.
28 g
 100 = 78% Fe
%Fe =
36 g
%O =
8.0 g
36 g
 100 = 22% O
A. Percentage Composition
 How
many grams of copper are in
a 38.0-gram sample of Cu2S?
Cu2S is 79.852% Cu
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
A. Percentage Composition
 Find
the mass percentage of
water in calcium chloride
dihydrate, CaCl2•2H2O?
%H2O =
36.04 g
 100 = 24.51%
H2O
147.02 g
B. Empirical Formula
 Smallest
whole number ratio of
atoms in a compound
C 2H 6
reduce subscripts
CH3
B. Empirical Formula Calculation
1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to
find subscripts.
4. When necessary, multiply
subscripts by 2, 3, or 4 to get
whole #’s.
B. Empirical Formula
 Find
the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
= 1.85 mol N
=1N
1.85 mol
14.01 g
74.1 g 1 mol
= 4.63 mol O
= 2.5 O
16.00 g
1.85 mol
B. Empirical Formula
N1O2.5
Need to make the subscripts whole
numbers  multiply by 2
N2O5
C. Molecular Formula
 “True Formula” - the actual number
of atoms in a compound
empirical
formula
CH3
?
molecular
formula
C2H6
C. Molecular Formula Calculation
1. Find the empirical formula.
2. Calculate the empirical formula
molar mass.
3. Divide the molecular molar mass by
the empirical molar mass.
4. Multiply each subscript by the
answer from step 3.
MF mass
n
EF mass
EF n
C. Molecular Formula
 The
empirical formula for ethylene
is CH2. Find the molecular formula
if the molecular mass is
28.06 g/mol?
empirical mass = 14.03 g/mol
28.06 g/mol
14.03 g/mol
= 2.00
(CH2)2  C2H4
Topic 9 The Mole
III. Molarity
I
II
III
IV
A. Molarity

Concentration of a solution.
substance being dissolved
moles of solute
Molarity (M) 
liters of solution
total combined volume
A. Molarity
2M HCl What does this mean?
mol
M
L
2 mol HCl
2M HCl 
1L
B. Molarity Calculations
molar mass
(g/mol)
6.02  1023
(particles/mol)
MASS
IN
NUMBER
MOLES
OF
GRAMS
PARTICLES
Molarity (mol/L)
Aqueous solution
LITERS
OF
SOLUTION
B. Molarity Calculations
 How
many grams of NaCl are
required to make 0.500L of 0.25M
NaCl?
0.500 L 0.25 mol 58.44 g
1L
0.25 mol
0.25M 
1L
1 mol
= 7.3 g NaCl
B. Molarity Calculations
 Find
the molarity of a 250 mL
solution containing 10.0 g of NaF.
10.0 g 1 mol
mol
M
L
M=
41.99 g
0.238 mol
0.25 L
= 0.238 mol NaF
= 0.95M NaF
C. Molar Volume at STP
1 mol of any gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm
Pressure
C. Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02 
MOLES
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
C. Molar Conversion Examples
 How
many Liters are in 2.50
moles of of Helium gas at STP?
2.50 mol 22.4 L
1 mol
= 56.0 L He
C. Molar Conversion Examples
 Find
the mass of 137 L of SO2
gas at STP.
137 L SO2
1 mol
64.07 g
22.4 L SO2 1 mol
= 392 g SO2
Find the number of Liters of CO2
at STP if it has a mass of 132
grams.